WileyPLUS W wileyplus.com WileyPLUS Fin ding electric potential, Physical Consta

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WileyPLUS W wileyplus.com WileyPLUS Fin ding electric potential, Physical Constants-The P + https://edugen.wileyplus.com/edugen/student/mainfr.uni Assienment>upen Assignment FULL SCREEN PRINTER VERSION BACK ASSIGNMENT RESOURCES Chapter 24, Problem 057 DiFabio Phys 202 Your answer is partially correct. Try again. uestioM ldentical 44 yC charges are fixed on an x axis at x = ±3.2 m. A particle of charge q =-19 s then released from rest at a point on the positive part of the y axis. Due to the symmetry of the situation, the particle moves along the y axis and has kinetic energy 1.0 J as it passes through the point x 0, y4.4 m. (a) What is the kinetic energy of the particle as it passes through the origin? (b) At what negative value of y will the particle momentarily stop? (a) Number 2.9366 014 (b) Numb Units 041 Click if you would like to Show Work for this question: Open Show Work Probl SHOW HINT 051 LINK TO TEX LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM VIDEO MINI-LECTURE By accessing this Question Asslstance, you wl learn whlle you earn points based on the Polnt Potential Policy set by your Instructor oterarnooien | Question Attempts: UnlimitedV OR LAIER SUBMII ANSWER 24.34 Earn Maximum Points available only if you answer this question correctly in four attempts or less. 034 All Rights Reserved. A Division ol Ju Version 4.24.5.1 - 11:50 AM Type here to search 3/15/2018

Explanation / Answer

(A) at given location,

V = k q1 / r1 + k q2 / r2

V = 2(9 x 10^9)(44 x 10^-6) / sqrt(3.2^2 + 4.4^2)

V = 1.46 x 10^5 Volt

at origin,

V = 2(9 x 10^9)(44 x 10^-6) / (3.2)

V = 2.47 x 10^5 Volt

PEi + KEi = PEf + KE f

(-19 x 10^-6)(1.46 x 10^5) + 1 = (-19 x 10^-6)(2.47 x 10^5) + KE

KE = 2.9 J  

(B) Pei + KEi = PEf + KEF

(-19 x 10^-6)(1.46 x 10^5) + 1 = PEf + 0

q Vf = - 1.774

Vf = 0.934 x 10^4 Volt

0.934 x 10^4 = 2(9 x 10^9)(44 x 10^-6) / sqrt(y^2 + 3.2^2)

y^2 + 3.2^2 = 71.95

y = - 7.9 m

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